THE INDUCTION MOTOR. 



73. In order to develop a maximum torque equal to 400 synchronous 

 horse-power, the maximum ordinate or the radius of the semi-circle 

 in the diagram must be equal to 



400 . 746 



= 102 amperes, 



1/3 . 2000 . 0.85 



assuming the efficiency to be 85 per cent at this output. Hence follows 

 that the diameter of the semi-circle must be equal to 200 amperes. 



74. The no-load current io can now immediately be calculated, it 

 we remember that the quotient of it and the diameter of the semi- 

 circle must be equal to <* = 0.061. We thus get 



t = 12 amperes. 



75. From formula (6) follows the number of conductors per pole and 

 phase, if a value for ($> is assumed. We make (B, the maximum induc- 

 tion in the air-gap, equal to 5600, and we have then 



*o = 12 amperes. 

 A = 0.15 cm. 

 (B = 5600 c. g. s. 



76. Inserting these values into the formula 



_ ffi . 1.6 A 



< 6) ' ~ 2 n V2 



we get 



n = 40 



77. The total number of active conductors per phase, z, is equal to n 

 multiplied by the number of poles. Thus we have 



z = 16 . 40 = 640 



78. We can now calculate the number of lines of induction per pole 

 with the help of formula (5). It is 



(5) e =2.12. ~. z . * . io- 8 



2OOO 



e = volts. 



V3 



~ =60 

 z = 640 

 Hence, 



$ = 1.42 . io' c. g. s. 



44 



